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 A292443 a(n) = (5/32)*A000045(6*n)^2. 1
 0, 10, 3240, 1043290, 335936160, 108170400250, 34830532944360, 11215323437683690, 3611299316401203840, 1162827164557749952810, 374426735688279083601000, 120564246064461307169569210, 38821312806020852629517684640, 12500342159292650085397524884890 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Every term is a triangular number. [Problem B-967 in Euler and Sadek, 2003; solution in Euler and Sadek, 2004] LINKS G. C. Greubel, Table of n, a(n) for n = 0..395 Russ Euler and Jawad Sadek, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 41, No. 5 (2003), 466-471. Russ Euler and Jawad Sadek, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 42, No. 3 (2004), 277-282. Index entries for linear recurrences with constant coefficients, signature (323,-323,1). FORMULA From Colin Barker, Sep 16 2017: (Start) G.f.: 10*x*(1 + x) / ((1 - x)*(1 - 322*x + x^2)). a(n) = ((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^n)^2) / 32. a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 2. (End) MATHEMATICA Table[(5/32) Fibonacci[6 n]^2, {n, 0, 13}] (* Michael De Vlieger, Sep 16 2017 *) PROG (PARI) a(n) = (5/32)*fibonacci(6*n)^2 (MAGMA) [5*Fibonacci(6*n)^2/32: n in [0..20]]; // G. C. Greubel, Feb 03 2019 (Sage) [5*fibonacci(6*n)^2/32 for n in (0..20)] # G. C. Greubel, Feb 03 2019 (GAP) List([0..20], n-> 5*Fibonacci(6*n)^2/32) # G. C. Greubel, Feb 03 2019 CROSSREFS Cf. A000045. Subsequence of A000217. Sequence in context: A225764 A243008 A133198 * A001329 A007101 A007103 Adjacent sequences:  A292440 A292441 A292442 * A292444 A292445 A292446 KEYWORD nonn,easy AUTHOR Felix FrÃ¶hlich, Sep 16 2017 STATUS approved

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Last modified February 20 22:47 EST 2020. Contains 332086 sequences. (Running on oeis4.)